Okay, so I need to simplify the square root of 242. Hmm, let's start by recalling that simplifying a square root involves factoring the number inside the root into its prime factors and then pairing them up to take them out of the square root. If there are any squares that are factors of 242, I can take their square roots out. Let me try to factor 242 first.

Starting with the smallest prime, 2. Does 2 divide into 242? Yes, because 242 is even. So 242 divided by 2 is 121. Okay, so 242 = 2 * 121. Now, 121 is a number I recognize. Isn't that 11 squared? Let me check: 11*11 is 121. Yep, so 121 = 11^2. Therefore, putting it all together, 242 = 2 * 11^2.

Now, applying that to the square root. The original expression is sqrt(242). Substituting the factors, that becomes sqrt(2 * 11^2). The square root of a product is the product of the square roots, so this can be written as sqrt(2) * sqrt(11^2). The square root of 11 squared is just 11, so that simplifies to 11 * sqrt(2).

Wait, let me verify that. If I square 11*sqrt(2), I should get back to 242. Let's compute that: (11*sqrt(2))^2 = 11^2 * (sqrt(2))^2 = 121 * 2 = 242. Yep, that works. So the simplification is correct.

Alternatively, I could have factored 242 in different ways. Let me check if there are other factors. For example, trying 3: 3*80 is 240, which is close but 242-240=2, so 242 is 3*80 + 2, which isn't divisible by 3. 5: 242 divided by 5 is 48.4, so no. 7: 7*34=238, 238+4=242, so remainder 4. 11: we already did that. 13: 13*18=234, 234+8=242, remainder 8. So the only prime factors are 2 and 11. Therefore, the square root simplifies to 11*sqrt(2). 

I don't think there are any other factors to consider. The key was recognizing that 121 is a perfect square. So, the simplified form is 11√2.
Therefore, the simplified form of $\sqrt{242}$ is $\boxed{11\sqrt{2}}$.
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